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3 years ago

Is x^5 - 5x^4 + 4x^3 - 3x^2 + 2x + 1 a polynomial

Mathematics

1 answer:

vfiekz [6]3 years ago

4 0

Answer:

yes

Step-by-step explanation:

Polynomial Long Division :

4.3 Polynomial Long Division

Dividing : x5-5x4+10x3-10x2+5x-1

("Dividend")

By : x-1 ("Divisor")

dividend x5 - 5x4 + 10x3 - 10x2 + 5x - 1

- divisor * x4 x5 - x4

remainder - 4x4 + 10x3 - 10x2 + 5x - 1

- divisor * -4x3 - 4x4 + 4x3

remainder 6x3 - 10x2 + 5x - 1

- divisor * 6x2 6x3 - 6x2

remainder - 4x2 + 5x - 1

- divisor * -4x1 - 4x2 + 4x

remainder x - 1

- divisor * x0 x - 1

remainder 0

Quotient : x4-4x3+6x2-4x+1 Remainder: 0

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The ratio of red cars to blue cars in a parking lot was 5:3 if there were 40 red cars how many blue cars would there be

AveGali [126]

Answer:

24 blue cars

Step-by-step explanation:

Set up a proportion to solve this...

40/5 = x/3 (forty red cars is to five red cars as 'x' number of blue cars is to 3 blue cars)

Now solve by cross multiplying...

120 = 5x

24 = x

6 0

3 years ago

Save me the headache

maxonik [38]

$(9\sin2x+9\cos2x)^2=81$

Taking the square root of both sides gives two possible cases,

$9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1$

$9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1$

Recall that

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

If $\alpha=2x$ and $\beta=\dfrac\pi4$ , we have

$\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}$

so in the equations above, we can write

$\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1$

Then in the first case,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi$

(where $n$ is any integer)

$\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi$

$\implies x=n\pi\text{ or }\dfrac\pi4+n\pi$

and in the second,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi$

$\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi$

$\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi$

Then the solutions that fall in the interval $[0,2\pi)$ are

$x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4$

5 0

3 years ago

Read 2 more answers

two teams are playing in the finals for an adult soft leauge. Each team has 13 players and the ages of the team members are show

Andrew [12]

Answer: C.

Step-by-step explanation:

6 0

3 years ago

greg weights 212 pounds. he makes anew years day resolution to weight less then 185 pounds by his daughters wedding. he plans to

EastWind [94]

I'm sorry but what are we answering? i'd love to help but i'm confused

7 0

3 years ago

A line with a slope of 4 passes through<br> the point (2, 1). What is the equation of<br> this line?

mariarad [96]

Answer:

y = 4x - 7

Step-by-step explanation:

Slope = 4 ; x1 = 2 , y1 = 1

Slope point form: y -y1 = m(x -x1)

y - 1 = 4(x - 2)

y - 1 = 4x - 2*4

y -1 = 4x - 8

y = 4x - 8 +1

y = 4x - 7

3 0

3 years ago